Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1000 + 17;
const long long INF = 1e9 + 17;
int n, a[MaxN][MaxN], m;
vector<pair<int, int> > Ans;
set<int> S;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) scanf("%d", a[i] + j);
  while (true) {
    int Min = INF;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
        if (a[i][j] != 0) Min = min(Min, a[i][j]);
    if (Min == INF) break;
    int Q = INF, posx, posy;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        if (a[i][j] == Min) {
          int Z = INF;
          for (int k = 1; k <= n; ++k) Z = min(Z, a[k][j]);
          for (int k = 1; k <= m; ++k) Z = min(Z, a[i][k]);
          if (Z < Q) {
            Q = Z;
            posx = i;
            posy = j;
          }
          if (Z == 0) break;
        }
      }
    }
    int W = INF, Z = INF;
    for (int k = 1; k <= n; ++k) Z = min(Z, a[k][posy]);
    for (int k = 1; k <= m; ++k) W = min(W, a[posx][k]);
    if (Z == 0 && W == 0) printf("-1"), exit(0);
    if (Z == 0 || W != 0 && n <= m) {
      Ans.push_back(make_pair(2, posx));
      for (int k = 1; k <= m; ++k) a[posx][k]--;
      --Z;
    } else {
      Ans.push_back(make_pair(1, posy));
      for (int k = 1; k <= n; ++k) a[k][posy]--;
    }
  }
  printf("%d\n", Ans.size());
  for (int i = 0; i < Ans.size(); ++i) {
    if (Ans[i].first == 1)
      printf("col %d\n", Ans[i].second);
    else
      printf("row %d\n", Ans[i].second);
  }
  return 0;
}